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lana [24]
3 years ago
12

Suppose that there are three astronauts outside a spaceship and they decide to play catch. all the astronauts weigh the same on

earth and are equally strong. the first astronaut throws the second one toward the third one and the game begins. describe the motion of the astronauts as the game proceeds. how long will the game last?
Mathematics
1 answer:
Lena [83]3 years ago
5 0
<span>The 3rd astronaut would catch the 2nd astronaut and throw the 2nd astronaut towards the 1st and the game would end there. The key thing to remember is conservation of momentum. Since all of the astronauts have the same mass and strength, I will be introducing a unit called "A" which represents the maximum momentum that one astronaut can produce while throwing another. So here's the game of catch, throw by throw. Before the game begins, I will assume all three astronauts are stationary and have 0 momentum. So Astronaut 1 = 0 A (Stationary, next to astronaut 2) Astronaut 2 = 0 A (Stationary, next to astronaut 1) Astronaut 3 = 0 A (Stationary) 1st astronaut grabs the 2nd astronaut and throws him towards the 3rd. Since every action has an equal and opposite reaction, what will happen is the 1st astronaut will be sent moving backwards with a momentum of -1/2A and the 2nd astronaut will be heading towards the 3rd with a momentum of +1/A. So we're left with Astronaut 1 = -1/2 A (Moving to the left) Astronaut 2 = +1/2 A (Moving to the right) Astronaut 3 = 0 A (Stationary) Now the 3rd astronaut catches the 2nd who was thrown at him. Both of them continue moving in the same direction as the 2nd astronaut was just prior to being caught, but at a reduced velocity, giving Astronaut 1 = -1/2 A (Moving to the left) Astronaut 2 = +1/4 A (Moving to the right, slowly) Astronaut 3 = +1/4 A (Moving to the right, slowly) Finally, Astronaut 3 throws astronaut 2 back towards Astronaut 1, giving Astronaut 1 = -1/2 A (Moving to the left) Astronaut 2 = +1/4 A -1/2A = -1/4A (Moving to the left, slowly) Astronaut 3 = +1/4 A +1/2A = +3/4A (Moving to the right, rapidly) So what you're left with is Astronaut 1 moving to the left faster than Astronaut 2, so those two astronauts will never catch each other. Meanwhile, Astronaut 3 is moving to the right and getting further and further away from the other 2 astronauts. So none of the astronauts will ever be able to catch or throw anyone ever again.</span>
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Ivahew [28]

In 20 minutes, John can wrap 10 small boxes and Hank can wrap 12 small boxes. So, Hank can wrap more small boxes in 20 minutes than John can.

Step-by-step explanation:

Given,

John wraps;

2 small boxes = 4 minutes

1 small box = \frac{4}{2} = 2\ minutes

In 20 minutes,

20 minutes = \frac{Total\ time}{Time per box} = \frac{20}{2}

20 minutes = 10 small boxes

Hank wraps;

3 small boxes = 5 minutes

1 small box =\frac{5}{3}\ minutes

In 20 minutes = \frac{Total\ time}{Time per box} = \frac{20}{5/3}

20\ minutes=\frac{20*3}{5}\\20\ minutes=\frac{60}{5}\\20\ minutes= 12\ small\ boxes\\

In 20 minutes, John can wrap 10 small boxes and Hank can wrap 12 small boxes. So, Hank can wrap more small boxes in 20 minutes than John can.

Keywords: division, multiplication

Learn more about division at:

  • brainly.com/question/1836777
  • brainly.com/question/2115122

#LearnwithBrainly

8 0
3 years ago
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Gre4nikov [31]
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sesenic [268]

Answer:

Step-by-step explanation:

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